now what you want to do is find the max value for x
so you do this by setting the dervative of V = 5x^2 -2x^3 to 0
then you get V'= 10x-6x^2 = 0
solve that and you get x=5/3m, or x is approx. 1.67m
and you can show that this is the max and no the min, by taking the second derivative, and since that answer will always be negative, therefore the graph in concave down, meaning its the mac value.
hopefully that made sence